ok well your answers to 1 and 3 are correct. however you can trivially see that 1/14.5 is lower than 1/13 (the chance you'll have an ace with 1 card), which can't possibly be right.

ANYWAY disregarding the exact math, the point is that the "implication" i suggested is NOT true, and it's not true in parallel question about coins and children either. The probabilities are different under different conditions.

Let me see if I understand what Aesah is saying about the coin flips. The way I first interpreted it was that hes going to flip a coin, if its heads hes going to flip a second coin and the odds of that one being heads is not 50/50, but that's not what hes saying, the odds are 50/50.

What I believe he is saying is that hes going to flip two coins and tell you if one is heads. In this case you can see that 75% of the time he will tell you one is heads. (HT TH HH TT). Then you have a 67% chance that the remaining coin is tails.

To piggyback off of sleepyboy, i think i truly understand what is going on here. When looking at the original post, some of us were looking at it differently than others. I and and the majority were looking at it from the vantage point of one coin has already flipped and now we are ready to flip the second one. This is why we all believed the answer to be 50% or 1/2 or whatever. ACK and aesah were looking at it from the view point that all of the coin flipping was done in the past and they were just looking at the data. From that POV, then yes the answer is 1/3 or 13/27 for Q3

ok well your answers to 1 and 3 are correct. however you can trivially see that 1/14.5 is lower than 1/13 (the chance you'll have an ace with 1 card), which can't possibly be right.

ANYWAY disregarding the exact math, the point is that the "implication" i suggested is NOT true, and it's not true in parallel question about coins and children either. The probabilities are different under different conditions.

#2 is astonishingly simple if you just think in reverse. What are the odds that neither of my two cards contains an Ace? A lot of probability and/or combinatorics problems are much easier solved by trying to solve for the opposing outcome.

48 out of 52 cards are not an Ace

48/52 - First card is not an Ace
48/51 - Second card is not an Ace, if First card is not an Ace
48/52 * 48/51 - First card is not an Ace AND Second Card is not an Ace

48/52 * 48/51 = 2304/2652 = approx. 86.9%

Simply subtract 0.869 from 1, and you have your answer: 13.1% OR 1/7.6

To piggyback off of sleepyboy, i think i truly understand what is going on here. When looking at the original post, some of us were looking at it differently than others. I and and the majority were looking at it from the vantage point of one coin has already flipped and now we are ready to flip the second one. This is why we all believed the answer to be 50% or 1/2 or whatever. ACK and aesah were looking at it from the view point that all of the coin flipping was done in the past and they were just looking at the data. From that POV, then yes the answer is 1/3 or 13/27 for Q3

This is the correct answer. The first thing you learn in probability is that you have to define exactly what you're trying to solve for. This whole thread was a lesson in semantics more than it was probability.